Answer:
The correct answer is: B. 14 H⁺ + 6 e⁻ + Cr₂O₇²⁻ ⟶ 2 Cr³⁺ + 7 H₂O
Explanation:
Redox reactions are reactions where oxidation and reduction processes happen at the same time through the simultaneous transfer of electrons between chemical species.
Reduction of a chemical species is illustrated in a reduction half- reaction, whereas oxidation is depicted in an oxidation half- reaction.
To balance the reduction half-reaction for converting Cr₂O₇²⁻ to Cr³⁺:
Cr₂O₇²⁻ ⟶ Cr³⁺
Initially, we balance the number of Cr atoms on both sides
Cr₂O₇²⁻ ⟶ 2 Cr³⁺
Here, Cr is in +6 oxidation state in Cr₂O₇²⁻, while it is +3 in Cr³⁺. Consequently, to get reduced, each Cr atom needs to gain 3 electrons.
Thus, 2 Cr atoms from Cr₂O₇²⁻ acquire a total of 6 electrons to reduce.
Cr₂O₇²⁻ + 6 e⁻ ⟶ 2 Cr³⁺
Next, the total charge on the reactant side is (-8), while on the product side it is (+6).
From the given options, it's clear that this reaction must be balanced under acidic conditions.
Thus, to equalize the overall charge on both sides,[ [TAG_58]]14 H⁺ ions are added to the reactant side.
Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺
To achieve balance among hydrogen and oxygen atoms, we add 7 H₂O to the product side.
Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O
Thus, the balanced reduction half-reaction is:
Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O
